Question

A small mirror of area 'A' and mass 'm' is suspended in a vertical plane by means of a weightless string. A beam of light of intensity 'I' falls normally on the mirror and the string is deflected form the vertical through a very small angle ' $\alpha$ '. Assuming the mirror to be perfectly reflecting, obtain an expression for $\alpha$

• A. $\frac{2 IA }{ c mg }$
• B. $\frac{I}{c m g}$
• C. $\frac{3 I}{c m g}$
• D. None of the above

Answer 1

Answer: (A)

When the light is incident on the mirror is totally reflected and there is no loss of energy.
$\therefore $ Change in momentum $=2 \times$ incident momentum
Intensity $ =\frac{\text { Energy }}{\text { Time } \times \text { Area }} $
Momentum $ =\frac{\text { Intensity }}{\text { Velocity of light }} $
Pressure, $P =\frac{\text { Force }}{\text { Area }}$
$=\frac{\text { Rate of change in momentum }}{ A } $
$= \frac{2IA}{c} $( where $c = $ speed of light)
$\alpha=\tan \alpha=\frac{ P \times A }{ mg }=\frac{2 IA }{ c mg }$